Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=505;
int read(){
	int f=1,g=0;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n;
ll pr[N+N],a[N][N],b[N+N];
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return b?(a/gcd(a,b)*b):a;}
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	scanf("%d",&n);n+=2;
	for (int i=2;pr[0]<=n+n+2;i++){
		int fl=1;
		for (int j=2;j*j<=i;j++)
		if (i%j==0){fl=0;break;}
		if (fl) pr[++pr[0]]=i;
	}
	int i=1,j=n+n+2;
	while (i<j){
		b[++b[0]]=pr[i++];
		b[++b[0]]=pr[j--];
	}
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
	if (!((i+j)&1)){
		int x=(i+j)>>1,y=(n-i+1+j)>>1;
		a[i][j]=b[x]*b[y+n];
	}
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
	if ((i+j)&1){
		a[i][j]=1;
		a[i][j]=lcm(a[i][j],a[i-1][j]);
		a[i][j]=lcm(a[i][j],a[i+1][j]);
		a[i][j]=lcm(a[i][j],a[i][j-1]);
		a[i][j]=lcm(a[i][j],a[i][j+1]);
		a[i][j]++;
	}
	for (int i=2;i<n;i++)
	for (int j=2;j<n;j++)
	printf("%lld%c",a[i][j],(j==n-1) ? '\n' : ' ');
	return 0;
}